Assignment 02 - Dselection_3_7 (Solution)

Assignment 02 - Dselection_3_7 (Solution) - The Hong Kong...

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Unformatted text preview: The Hong Kong University of Science & Technology COMP 271: Design and Analysis of Algorithms Fall 2007 Solution to Assignment 2 In the DSelection algorithm described in Lecture 5, input elements are di- vided into groups of 5. The running time is then O ( n ) in the worst case. In this assignment, we ponder what if the elements are divided into groups of 3 or 7. For convenience of exposition, we index the array from 1 to n as shown in the following figure. 1 n Part 1: When Each Group Contains 7 Elements ( 5 points) DSelection has five steps. Refer to lecture notes. 1. Divide the n items into ⌈ n/ 7 ⌉ groups in which each, except possibly the last, contains 7 elements. O ( n ) 2. Find the median of each of the ⌈ n/ 7 ⌉ groups. O ( n ) 3. Take these ⌈ n/ 7 ⌉ medians and put them in another array. Use DSelection() to recursively calculate the median of these me- dians. Call this x . T ( ⌈ n/ 7 ⌉ ) 4. Partition the original array using x as the pivot. 1 k x ↑ k-th element n Suppose x is the k-th element in the array. It is the k-th small- est element as well. 5. If i = k return x if i < k return DSelection ( A, 1 ,k − 1 ,i ) if i > k return DSelection ( A,k +1 ,n,i ) 1 Denote by T ( n ) the running time in the worst case. We can easily set up the recurrence inequality: T ( n ) ≤ T ( ⌈ n/ 7 ⌉ ) + T (max { k − 1 ,n − k } ) + an (1) In lecture, we have a lemma which claims that the pivot x is “near” the center of the array. A similar result holds for the case of groupthe center of the array....
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This note was uploaded on 10/18/2009 for the course COMP 271 taught by Professor Arya during the Spring '07 term at HKUST.

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Assignment 02 - Dselection_3_7 (Solution) - The Hong Kong...

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